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Algebra of Derivatives

Prove: 1+cot...

Question

Prove:
$(1 + c o t A + t a n A) (s i n A - c o s A) = s i n A t a n A - c o t A c o s A$

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Solution

$\begin{matrix} L . H . S = (1 + cot A + tan A) (sin A - cos A) = (sin A + sin A cot A + tan A sin A - cos A - cos A cot A - tan A cos A) = (sin A + \frac{s i n A c o s A}{sin A} + tan A sin A - cos A - cos A cot A - \frac{cos A sin A}{cos A}) = (sin A + cos A + tan A sin A - cos A - cos A cot A - sin A) = tan A sin A - c o s A c o t A = s i n A t a n A - c o t A c o s A = R . H . S P r o v e d . \end{matrix}$

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(1 + c o t A + t a n A) (s i n A - c o s A)

\frac{s e c A}{c o s e c^{2} A} - \frac{c o s e c A}{s e c^{2} A}

s i n A t a n A - c o t A c o s A

(1 + cot A + tan A) (sin A - cos A) = sin A tan A - cot A cos A

Q. Prove the following trigonometric identities.

(1 + \cot A + \tan A) (\sin A - \cos A) = \frac{\sec A}{{cosec}^{2} A} - \frac{cosec A}{\sec^{2} A} = \sin A \tan A - \cot A \cos A

Q. (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A

Q. Prove that :

\frac{(1 + c o t A + t a n A) (s i n a - c o s A)}{s e c^{3} A - c o s e c^{3} A} = s i n^{2} A c o s^{2} A

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